cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 27 results. Next

A290149 Totient sublime numbers: numbers k such that the number of terms in the iterations of phi(k) from k to 1, A032358(k)+2, and their sum, A092693(k) are both perfect totient numbers (A082897).

Original entry on oeis.org

6, 2916, 4374, 109100, 113708, 3188646
Offset: 1

Views

Author

Amiram Eldar, Jul 21 2017

Keywords

Comments

Analogous to A081357 (sublime numbers), as A082897 (perfect totient numbers) is analogous to A000396 (perfect numbers).
No other terms below 10^8.

Examples

			There are 9 terms in the iterations of phi(k) for 2916: 2916, 972, 324, 108, 36, 12, 4, 2, 1. Their sum is 4375. Both 9 and 4375 are perfect totient numbers (A082897).
		

Crossrefs

Programs

  • Mathematica
    iterList [n_] := FixedPointList[EulerPhi@# &, n]; sumIter [n_] := Plus @@ iterList[n] - 1; numIter[n_] := Length[iterList[n]] - 1; perfTotQ[n_] := sumIter[n] == 2 n; totSublimeQ[n_] := perfTotQ[numIter[n]] && perfTotQ[sumIter[n]]; Select[Range [10^8], totSublimeQ]

A000244 Powers of 3: a(n) = 3^n.

Original entry on oeis.org

1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987
Offset: 0

Views

Author

Keywords

Comments

Same as Pisot sequences E(1, 3), L(1, 3), P(1, 3), T(1, 3). Essentially same as Pisot sequences E(3, 9), L(3, 9), P(3, 9), T(3, 9). See A008776 for definitions of Pisot sequences.
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2n + 2, s(0) = 1, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004
a(1) = 1, a(n+1) is the least number such that there are a(n) even numbers between a(n) and a(n+1). Generalization for the sequence of powers of k: 1, k, k^2, k^3, k^4, ... There are a(n) multiples of k-1 between a(n) and a(n+1). - Amarnath Murthy, Nov 28 2004
a(n) = sum of (n+1)-th row in Triangle A105728. - Reinhard Zumkeller, Apr 18 2005
With p(n) being the number of integer partitions of n, p(i) being the number of parts of the i-th partition of n, d(i) being the number of different parts of the i-th partition of n, m(i, j) being the multiplicity of the j-th part of the i-th partition of n, Sum_{i = 1..p(n)} being the sum over i and Product_{j = 1..d(i)} being the product over j, one has: a(n) = Sum_{i = 1..p(n)} (p(i)!/(Product_{j = 1..d(i)} m(i, j)!))*2^(p(i) - 1). - Thomas Wieder, May 18 2005
For any k > 1 in the sequence, k is the first prime power appearing in the prime decomposition of repunit R_k, i.e., of A002275(k). - Lekraj Beedassy, Apr 24 2006
a(n-1) is the number of compositions of compositions. In general, (k+1)^(n-1) is the number of k-levels nested compositions (e.g., 4^(n-1) is the number of compositions of compositions of compositions, etc.). Each of the n - 1 spaces between elements can be a break for one of the k levels, or not a break at all. - Franklin T. Adams-Watters, Dec 06 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then a(n) = |S|. - Ross La Haye, Dec 22 2006
From Manfred Boergens, Mar 28 2023: (Start)
With regard to the comment by Ross La Haye:
Cf. A001047 if either nonempty subsets are considered or x is a proper subset of y.
Cf. a(n+1) in A028243 if nonempty subsets are considered and x is a proper subset of y. (End)
If X_1, X_2, ..., X_n is a partition of the set {1, 2, ..., 2*n} into blocks of size 2 then, for n >= 1, a(n) is equal to the number of functions f : {1, 2, ..., 2*n} -> {1, 2} such that for fixed y_1, y_2, ..., y_n in {1, 2} we have f(X_i) <> {y_i}, (i = 1, 2, ..., n). - Milan Janjic, May 24 2007
This is a general comment on all sequences of the form a(n) = [(2^k)-1]^n for all positive integers k. Example 1.1.16 of Stanley's "Enumerative Combinatorics" offers a slightly different version. a(n) in the number of functions f:[n] into P([k]) - {}. a(n) is also the number of functions f:[k] into P([n]) such that the generalized intersection of f(i) for all i in [k] is the empty set. Where [n] = {1, 2, ..., n}, P([n]) is the power set of [n] and {} is the empty set. - Geoffrey Critzer, Feb 28 2009
a(n) = A064614(A000079(n)) and A064614(m)A000079(n). - Reinhard Zumkeller, Feb 08 2010
3^(n+1) = (1, 2, 2, 2, ...) dot (1, 1, 3, 9, ..., 3^n); e.g., 3^3 = 27 = (1, 2, 2, 2) dot (1, 1, 3, 9) = (1 + 2 + 6 + 18). - Gary W. Adamson, May 17 2010
a(n) is the number of generalized compositions of n when there are 3*2^i different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
For n >= 1, a(n-1) is the number of generalized compositions of n when there are 2^(i-1) different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
The sequence in question ("Powers of 3") also describes the number of moves of the k-th disk solving the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] pre-colored Magnetic Tower of Hanoi puzzle (cf. A183111 - A183125).
a(n) is the number of Stern polynomials of degree n. See A057526. - T. D. Noe, Mar 01 2011
Positions of records in the number of odd prime factors, A087436. - Juri-Stepan Gerasimov, Mar 17 2011
Sum of coefficients of the expansion of (1+x+x^2)^n. - Adi Dani, Jun 21 2011
a(n) is the number of compositions of n elements among {0, 1, 2}; e.g., a(2) = 9 since there are the 9 compositions 0 + 0, 0 + 1, 1 + 0, 0 + 2, 1 + 1, 2 + 0, 1 + 2, 2 + 1, and 2 + 2. [From Adi Dani, Jun 21 2011; modified by editors.]
Except the first two terms, these are odd numbers n such that no x with 2 <= x <= n - 2 satisfy x^(n-1) == 1 (mod n). - Arkadiusz Wesolowski, Jul 03 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 3-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Explanation from David Applegate, Feb 20 2017: (Start)
Since the preceding comment appears in a large number of sequences, it might be worth adding a proof.
The number of compositions of n into exactly k parts is binomial(n-1,k-1).
For a p-colored composition of n such that no adjacent parts have the same color, there are exactly p choices for the color of the first part, and p-1 choices for the color of each additional part (any color other than the color of the previous one). So, for a partition into k parts, there are p (p-1)^(k-1) valid colorings.
Thus the number of p-colored compositions of n into exactly k parts such that no adjacent parts have the same color is binomial(n-1,k-1) p (p-1)^(k-1).
The total number of p-colored compositions of n such that no adjacent parts have the same color is then
Sum_{k=1..n} binomial(n-1,k-1) * p * (p-1)^(k-1) = p^n.
To see this, note that the binomial expansion of ((p - 1) + 1)^(n - 1) = Sum_{k = 0..n - 1} binomial(n - 1, k) (p - 1)^k 1^(n - 1 - k) = Sum_{k = 1..n} binomial(n - 1, k - 1) (p - 1)^(k - 1).
(End)
Also, first and least element of the matrix [1, sqrt(2); sqrt(2), 2]^(n+1). - M. F. Hasler, Nov 25 2011
One-half of the row sums of the triangular version of A035002. - J. M. Bergot, Jun 10 2013
Form an array with m(0,n) = m(n,0) = 2^n; m(i,j) equals the sum of the terms to the left of m(i,j) and the sum of the terms above m(i,j), which is m(i,j) = Sum_{k=0..j-1} m(i,k) + Sum_{k=0..i-1} m(k,j). The sum of the terms in antidiagonal(n+1) = 4*a(n). - J. M. Bergot, Jul 10 2013
a(n) = A007051(n+1) - A007051(n), and A007051 are the antidiagonal sums of an array defined by m(0,k) = 1 and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to left of m(n, k) plus those above m(n, k). m(1, k) = A000079(k); m(2, k) = A045623(k + 1); m(k + 1, k) = A084771(k). - J. M. Bergot, Jul 16 2013
Define an array to have m(0,k) = 2^k and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to the left of m(n, k) plus those above m(n, k). Row n = 0 of the array comprises A000079, column k = 0 comprises A011782, row n = 1 comprises A001792. Antidiagonal sums of the array are a(n): 1 = 3^0, 1 + 2 = 3^1, 2 + 3 + 4 = 3^2, 4 + 7 + 8 + 8 = 3^3. - J. M. Bergot, Aug 02 2013
The sequence with interspersed zeros and o.g.f. x/(1 - 3*x^2), A(2*k) = 0, A(2*k + 1) = 3^k = a(k), k >= 0, can be called hexagon numbers. This is because the algebraic number rho(6) = 2*cos(Pi/6) = sqrt(3) of degree 2, with minimal polynomial C(6, x) = x^2 - 3 (see A187360, n = 6), is the length ratio of the smaller diagonal and the side in the hexagon. Hence rho(6)^n = A(n-1)*1 + A(n)*rho(6), in the power basis of the quadratic number field Q(rho(6)). One needs also A(-1) = 1. See also a Dec 02 2010 comment and the P. Steinbach reference given in A049310. - Wolfdieter Lang, Oct 02 2013
Numbers k such that sigma(3k) = 3k + sigma(k). - Jahangeer Kholdi, Nov 23 2013
All powers of 3 are perfect totient numbers (A082897), since phi(3^n) = 2 * 3^(n - 1) for n > 0, and thus Sum_{i = 0..n} phi(3^i) = 3^n. - Alonso del Arte, Apr 20 2014
The least number k > 0 such that 3^k ends in n consecutive decreasing digits is a 3-term sequence given by {1, 13, 93}. The consecutive increasing digits are {3, 23, 123}. There are 100 different 3-digit endings for 3^k. There are no k-values such that 3^k ends in '012', '234', '345', '456', '567', '678', or '789'. The k-values for which 3^k ends in '123' are given by 93 mod 100. For k = 93 + 100*x, the digit immediately before the run of '123' is {9, 5, 1, 7, 3, 9, 5, 1, 3, 7, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus we see the digit before '123' will never be a 0. So there are no further terms. - Derek Orr, Jul 03 2014
All elements of A^n where A = (1, 1, 1; 1, 1, 1; 1, 1, 1). - David Neil McGrath, Jul 23 2014
Counts all walks of length n (open or closed) on the vertices of a triangle containing a loop at each vertex starting from any given vertex. - David Neil McGrath, Oct 03 2014
a(n) counts walks (closed) on the graph G(1-vertex;1-loop,1-loop,1-loop). - David Neil McGrath, Dec 11 2014
2*a(n-2) counts all permutations of a solitary closed walk of length (n) from the vertex of a triangle that contains 2 loops on each of the remaining vertices. In addition, C(m,k)=2*(2^m)*B(m+k-2,m) counts permutations of walks that contain (m) loops and (k) arcs. - David Neil McGrath, Dec 11 2014
a(n) is the sum of the coefficients of the n-th layer of Pascal's pyramid (a.k.a., Pascal's tetrahedron - see A046816). - Bob Selcoe, Apr 02 2016
Numbers n such that the trinomial x^(2*n) + x^n + 1 is irreducible over GF(2). Of these only the trinomial for n=1 is primitive. - Joerg Arndt, May 16 2016
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
a(n-1) is also the number of compositions of n if the parts can be runs of any length from 1 to n, and can contain any integers from 1 to n. - Gregory L. Simay, May 26 2017
Also the number of independent vertex sets and vertex covers in the n-ladder rung graph n P_2. - Eric W. Weisstein, Sep 21 2017
Also the number of (not necessarily maximal) cliques in the n-cocktail party graph. - Eric W. Weisstein, Nov 29 2017
a(n-1) is the number of 2-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
a(n) is the number of faces of any dimension (vertices, edges, square faces, etc.) of the n-dimensional hypercube. For example, the 0-dimensional hypercube is a point, and its only face is itself. The 1-dimensional hypercube is a line, which has two vertices and an edge. The 2-dimensional hypercube is a square, which has four vertices, four edges, and a square face. - Kevin Long, Mar 14 2023
Number of pairs (A,B) of subsets of M={1,2,...,n} with union(A,B)=M. For nonempty subsets cf. A058481. - Manfred Boergens, Mar 28 2023
From Jianing Song, Sep 27 2023: (Start)
a(n) is the number of disjunctive clauses of n variables up to equivalence. A disjunctive clause is a propositional formula of the form l_1 OR ... OR l_m, where l_1, ..., l_m are distinct elements in {x_1, ..., x_n, NOT x_1, ..., NOT x_n} for n variables x_1, ... x_n, and no x_i and NOT x_i appear at the same time. For each 1 <= i <= n, we can have neither of x_i or NOT x_i, only x_i or only NOT x_i appearing in a disjunctive clause, so the number of such clauses is 3^n. Viewing the propositional formulas of n variables as functions {0,1}^n -> {0,1}, a disjunctive clause corresponds to a function f such that the inverse image of 0 is of the form A_1 X ... X A_n, where A_i is nonempty for all 1 <= i <= n. Since each A_i has 3 choices ({0}, {1} or {0,1}), we also find that the number of disjunctive clauses of n variables is 3^n.
Equivalently, a(n) is the number of conjunctive clauses of n variables. (End)
The finite subsequence a(2), a(3), a(4), a(5) = 9, 27, 81, 243 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A007283 (see comment there). - Felix Huber, Feb 15 2024

Examples

			G.f. = 1 + 3*x + 9*x^2 + 27*x^3 + 81*x^4 + 243*x^5 + 729*x^6 + 2187*x^7 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A008776 (2*a(n), and first differences).
a(n) = A092477(n, 2) for n > 0.
a(n) = A159991(n) / A009964(n).
Cf. A100772, A035002. Row sums of A125076 and A153279.
a(n) = A217764(0, n).
Cf. A046816, A006521, A014945, A275414 (multisets).
The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).

Programs

Formula

a(n) = 3^n.
a(0) = 1; a(n) = 3*a(n-1).
G.f.: 1/(1-3*x).
E.g.f.: exp(3*x).
a(n) = n!*Sum_{i + j + k = n, i, j, k >= 0} 1/(i!*j!*k!). - Benoit Cloitre, Nov 01 2002
a(n) = Sum_{k = 0..n} 2^k*binomial(n, k), binomial transform of A000079.
a(n) = A090888(n, 2). - Ross La Haye, Sep 21 2004
a(n) = 2^(2n) - A005061(n). - Ross La Haye, Sep 10 2005
a(n) = A112626(n, 0). - Ross La Haye, Jan 11 2006
Hankel transform of A007854. - Philippe Deléham, Nov 26 2006
a(n) = 2*StirlingS2(n+1,3) + StirlingS2(n+2,2) = 2*(StirlingS2(n+1,3) + StirlingS2(n+1,2)) + 1. - Ross La Haye, Jun 26 2008
a(n) = 2*StirlingS2(n+1, 3) + StirlingS2(n+2, 2) = 2*(StirlingS2(n+1, 3) + StirlingS2(n+1, 2)) + 1. - Ross La Haye, Jun 09 2008
Sum_{n >= 0} 1/a(n) = 3/2. - Gary W. Adamson, Aug 29 2008
If p(i) = Fibonacci(2i-2) and if A is the Hessenberg matrix of order n defined by A(i, j) = p(j-i+1), (i <= j), A(i, j) = -1, (i = j+1), and A(i, j) = 0 otherwise, then, for n >= 1, a(n-1) = det A. - Milan Janjic, May 08 2010
G.f. A(x) = M(x)/(1-M(x))^2, M(x) - o.g.f for Motzkin numbers (A001006). - Vladimir Kruchinin, Aug 18 2010
a(n) = A133494(n+1). - Arkadiusz Wesolowski, Jul 27 2011
2/3 + 3/3^2 + 2/3^3 + 3/3^4 + 2/3^5 + ... = 9/8. [Jolley, Summation of Series, Dover, 1961]
a(n) = Sum_{k=0..n} A207543(n,k)*4^(n-k). - Philippe Deléham, Feb 25 2012
a(n) = Sum_{k=0..n} A125185(n,k). - Philippe Deléham, Feb 26 2012
Sum_{n > 0} Mobius(n)/a(n) = 0.181995386702633887827... (see A238271). - Alonso del Arte, Aug 09 2012. See also the sodium 3s orbital energy in table V of J. Chem. Phys. 53 (1970) 348.
a(n) = (tan(Pi/3))^(2*n). - Bernard Schott, May 06 2022
a(n-1) = binomial(2*n-1, n) + Sum_{k >= 1} binomial(2*n, n+3*k)*(-1)^k. - Greg Dresden, Oct 14 2022
G.f.: Sum_{k >= 0} x^k/(1-2*x)^(k+1). - Kevin Long, Mar 14 2023

A092693 Sum of iterated phi(n).

Original entry on oeis.org

0, 1, 3, 3, 7, 3, 9, 7, 9, 7, 17, 7, 19, 9, 15, 15, 31, 9, 27, 15, 19, 17, 39, 15, 35, 19, 27, 19, 47, 15, 45, 31, 35, 31, 39, 19, 55, 27, 39, 31, 71, 19, 61, 35, 39, 39, 85, 31, 61, 35, 63, 39, 91, 27, 71, 39, 55, 47, 105, 31, 91, 45, 55, 63, 79, 35, 101, 63, 79, 39, 109, 39, 111
Offset: 1

Views

Author

T. D. Noe, Mar 04 2004

Keywords

Comments

Iannucci, Moujie and Cohen examine perfect totient numbers: n such that a(n) = n.

Examples

			a(100) = 71 because the iterations of phi (40, 16, 8, 4, 2, 1) sum to 71.
		

Crossrefs

Cf. A003434 (iterations of phi(n) needed to reach 1), A092694 (iterated phi product).
Cf. A082897 and A091847 (perfect totient numbers).

Programs

  • Haskell
    a092693 1 = 0
    a092693 n = (+ 1) $ sum $ takeWhile (/= 1) $ iterate a000010 $ a000010 n
    -- Reinhard Zumkeller, Oct 27 2011
    
  • Mathematica
    nMax=100; a=Table[0, {nMax}]; Do[e=EulerPhi[n]; a[[n]]=e+a[[e]], {n, 2, nMax}]; a (* T. D. Noe *)
    Table[Plus @@ FixedPointList[EulerPhi, n] - (n + 1), {n, 72}] (* Alonso del Arte, Jan 29 2007 *)
  • PARI
    a(n)=my(k);while(n>1,k+=n=eulerphi(n));k \\ Charles R Greathouse IV, Mar 22 2012
    
  • Python
    from sympy import totient
    from math import prod
    def f(n):
        m = n
        while m > 1:
            m = totient(m)
            yield m
    def A092693(n): return sum(f(n)) # Chai Wah Wu, Nov 14 2021

Formula

a(1) = 0, a(n) = phi(n) + a(phi(n))
a(n) = A053478(n) - n. - Vladeta Jovovic, Jul 02 2004
Erdős & Subbarao prove that a(n) ~ phi(n) for almost all n. In particular, a(n) < n for almost all n. The proportion of numbers up to N for which a(n) > n is at most 1/log log log log N. - Charles R Greathouse IV, Mar 22 2012

A082894 a(n) is the closest number to 2^n which is divisible by n.

Original entry on oeis.org

2, 4, 9, 16, 30, 66, 126, 256, 513, 1020, 2046, 4092, 8190, 16380, 32775, 65536, 131070, 262152, 524286, 1048580, 2097144, 4194300, 8388606, 16777224, 33554425, 67108860, 134217729, 268435468, 536870910, 1073741820, 2147483646
Offset: 1

Views

Author

Labos Elemer, Apr 22 2003

Keywords

Examples

			n=11: 2^11=2048 is between 2046=11.186 and 2035=11.185, closer to a(11)=2046;
Powers of two are fixed points of this map.
		

Crossrefs

Programs

  • Maple
    A082894:=n->n*floor((floor(n/2)+2^n)/n); seq(A082894(k), k=1..100); # Wesley Ivan Hurt, Oct 29 2013
  • Mathematica
    Table[n*Floor[(Floor[n/2]+2^n)/n], {n, 100}]
  • PARI
    for(n=1,50, print1(n*floor( (floor(n/2)+2^n) / n ), ", ")) \\ G. C. Greubel, Aug 08 2017
    
  • Python
    def A082894(n): return (m:=(1<>1))-m%n # Chai Wah Wu, Apr 23 2025

Formula

a(n) = n*floor( (floor(n/2)+2^n) / n ).

A137815 Year numbers: numbers n such that phi(n) = 2 phi(sigma(n)).

Original entry on oeis.org

5, 13, 37, 61, 65, 73, 119, 157, 185, 193, 277, 305, 313, 365, 397, 421, 457, 481, 541, 613, 661, 673, 733, 757, 785, 793, 877, 949, 965, 997, 1093, 1153, 1201, 1213, 1237, 1321, 1381, 1385, 1453, 1547, 1565, 1615, 1621, 1657, 1753, 1873, 1933, 1985, 1993
Offset: 1

Views

Author

R. K. Guy, R. J. Mathar and M. F. Hasler, Feb 11 2008

Keywords

Comments

Following D. Iannucci, n is called a "year number" if phi(n) / phi(sigma(n)) = 2 (thus 365 is a year number, explaining the terminology).
D. Iannucci asks: Are there any even year numbers? Are there any odd year numbers that are not squarefree?
Remark: If n = q_1 q_2 ... q_k is a product of odd primes such that (q_j + 1)/2 is an odd prime for all j, then n is a year number.
Solution: for nonsquarefree year numbers, see A137816. See A137817-A137819 for year numbers with cubes, 4th powers, 5th powers.
Eric Landquist found year numbers divisible by 7^2, 7^3 and 7^4, as well as 120781449 = 3^8 * 41 * 449.
The existence of even year numbers is still open, but Eric checked all 200-smooth even integers with a single large prime up to 10^8 and found no year numbers among them.
See also references in A082897 (perfect totient numbers).

References

  • R. K. Guy, "Euler's Totient Function", "Solutions of phi(m)=sigma(n)", "Iterations of phi and sigma", "Behavior of phi(sigma(n)) and sigma(phi(n))". =A7 B36-B42 in Unsolved Problems in Number Theory, 3rd ed. New York: Springer-Verlag, pp. 138-151, 2004.
  • Doug Iannucci, in: Gerry Myerson (ed.), 2007 Western Number Theory problems set.

Crossrefs

Cf. A137816-A137819, A006872 (phi(sigma(n)) = phi(n)), A067704 (phi(sigma(n)) = 2 phi(n)), A082897.

Programs

  • Mathematica
    Select[Range[2000],EulerPhi[#]==2EulerPhi[DivisorSigma[1,#]]&]  (* Harvey P. Dale, Mar 18 2011 *)
  • PARI
    for( n=1,10^7, eulerphi(n)==2*eulerphi(sigma(n)) && print1(n", "))

A286265 Totient abundant numbers: numbers k such that A092693(k) > k.

Original entry on oeis.org

5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 33, 35, 37, 41, 43, 47, 49, 51, 53, 55, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 123, 125, 127, 131, 133, 137, 139, 141, 143, 145, 149, 151, 153, 155, 157
Offset: 1

Views

Author

Amiram Eldar, May 05 2017

Keywords

Examples

			19 is a totient abundant number since A092693(19) = phi(19) + phi(phi(19)) + ... = 18 + 6 + 2 + 1 = 27 > 19.
		

Crossrefs

Programs

  • Mathematica
    totAbundantQ[n_] := Plus @@ FixedPointList[ EulerPhi@ # &, n] > 2*n+1; Select[Range[1000],totAbundantQ]
  • Python
    from sympy import totient
    def a092693(n): return 0 if n==1 else totient(n) + a092693(totient(n))
    print([n for n in range(1, 201) if a092693(n) > n]) # Indranil Ghosh, May 05 2017

A286067 Unitary perfect totient numbers: numbers n that equal to the sum of their iterated unitary totient uphi(n).

Original entry on oeis.org

3, 10, 21, 110, 3910, 1500988838
Offset: 1

Views

Author

Amiram Eldar, May 01 2017

Keywords

Comments

The unitary version of A082897 (perfect totient numbers), in which the unitary totient function uphi(n) (A047994) replaces the Euler totient function phi(n) (A000010).
a(7) > 5*10^9, if it exists. - Giovanni Resta, May 06 2020

Examples

			3910 is a unitary perfect totient number because 3910 = uphi(3910) + uphi(uphi(3910)) + uphi(uphi(uphi(3910))) + ... = 1408 + 1270 + 504 + 336 + 180 + 96 + 62 + 30 + 8 + 7 + 6 + 2 + 1.
		

Crossrefs

Programs

  • Mathematica
    uphi[n_] := (Times @@ (Table[#[[1]]^#[[2]] - 1, {1}] & /@ FactorInteger[n]))[[1]]; kmax = 10000; a = Table[0, {kmax}]; uptns = {}; Do[e = uphi[k]; a[[k]] = e + a[[e]]; If[k == a[[k]], AppendTo[uptns , k]], {k, 2, kmax}]; uptns

Extensions

a(6) from Giovanni Resta, May 06 2020

A330273 Infinitary perfect totient numbers: numbers that equal to the sum of their iterated infinitary totient function (A091732).

Original entry on oeis.org

3, 10, 21, 44, 93, 118, 170, 320, 548, 3596, 3620, 4772, 5564, 18260, 33051, 256425, 403700, 1071129, 1790160, 2318180, 3968852, 4027375, 10001319, 11270012, 12048740, 13358121, 31741593, 46271673, 56149161, 4344134553
Offset: 1

Views

Author

Amiram Eldar, Dec 13 2019

Keywords

Comments

The infinitary version of A082897 (perfect totient numbers), in which the infinitary totient function iphi (A091732) replaces the Euler totient function (A000010).

Examples

			10 is an infinitary perfect totient number because iphi(10) + iphi(iphi(10)) + ... = 4 + 3 + 2 + 1 = 10.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(2^(-1 + Position[Reverse @ IntegerDigits[e, 2], ?(# == 1 &)])); iphi[1] = 1; iphi[n] := iphi[n] = Times @@ (Flatten@(f @@@ FactorInteger[n]) - 1); infPerfTotQ[n_] := Plus @@ FixedPointList[iphi@# &, n] == 2 n + 1; Select[Range[1000], infPerfTotQ]

A288452 Pseudoperfect totient numbers: numbers n such that equal the sum of a subset of their iterated phi(n).

Original entry on oeis.org

3, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 47, 49, 51, 53, 55, 59, 61, 65, 67, 69, 71, 73, 77, 79, 81, 83, 85, 87, 89, 97, 101, 103, 107, 109, 111, 113, 115, 119, 121, 123, 125, 127, 131, 137, 139, 141, 143, 149, 151, 153, 155
Offset: 1

Views

Author

Amiram Eldar, Jun 09 2017

Keywords

Comments

Analogous to A005835 (pseudoperfect numbers) as A082897 (perfect totient numbers) is analogous to A000396 (perfect numbers).
All the odd primes are in this sequence.
Number of terms < 10^k: 4, 40, 350, 2956, 24842, etc. - Robert G. Wilson v, Jun 17 2017
All terms are odd. If n is even, phi(n) <= n/2, and except for n = 2, we will have phi(n) also even. So the sum of the phi sequence < n*(1/2 + 1/4 + ...) = n. - Franklin T. Adams-Watters, Jun 25 2017

Examples

			The iterated phi of 25 are 20, 8, 4, 2, 1 and 25 = 20 + 4 + 1.
		

Crossrefs

Supersequence of A082897. Subsequence of A286265.

Programs

  • Mathematica
    pseudoPerfectTotQ[n_]:= Module[{tots = Most[Rest[FixedPointList[EulerPhi@# &, n]]]}, MemberQ[Total /@ Subsets[tots, Length[tots]], n]]; Select[Range[155], pseudoPerfectTotQ]
  • PARI
    subsetSum(v, target)=if(setsearch(v,target), return(1)); if(#v<2, return(target==0)); my(u=v[1..#v-1]); if(target>v[#v] && subsetSum(u, target-v[#v]), return(1)); subsetSum(u,target);
    is(n)=if(isprime(n), return(n>2)); my(v=List(),k=n); while(k>1, listput(v,k=eulerphi(k))); subsetSum(Set(v),n) \\ Charles R Greathouse IV, Jun 25 2017

A091847 Perfect totient numbers, omitting powers of 3.

Original entry on oeis.org

15, 39, 111, 183, 255, 327, 363, 471, 2199, 3063, 4359, 4375, 5571, 8751, 15723, 36759, 46791, 65535, 140103, 208191, 441027, 4190263, 9056583, 57395631, 172186887, 236923383, 918330183, 3932935775, 4294967295, 4764161215
Offset: 1

Views

Author

N. J. A. Sloane, Mar 13 2004

Keywords

Crossrefs

A082897 has more information.

Programs

  • Mathematica
    fQ[n_] := !IntegerQ@ Log[3, n] && Plus @@ FixedPointList[ EulerPhi@# &, n] == 2n + 1 (* Robert G. Wilson v, Nov 06 2010 *)
  • Python
    from itertools import count, islice
    from gmpy2 import digits
    from sympy import totient
    def A091847_gen(startvalue=3): # generator of terms >= startvalue
        for n in count((k:=max(startvalue,3))+1-(k&1),2):
            t = digits(n,3)
            if t.count('0') != len(t)-1:
                m, s = n, 1
                while (m:=totient(m))>1:
                    s += m
                if s == n:
                    yield n
    A091847_list = list(islice(A091847_gen(),10)) # Chai Wah Wu, Mar 24 2023
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